Question

$displaystyle text{arcctg};left(tanleft(frac{27pi}{7}right)right);\ arctanleft(text{ctg}left(frac{-21pi}{5}right)right)$

Answer 1

$1)qquad arcctg(ctgx)=x; ,; ; esli; ; -frac{pi}{2} textless x textless frac{pi}{2}\\arcctg(tgfrac{27pi}{7})=arcctg(tg(3pi +frac{6pi}{7}))=arcctg(tgfrac{6pi}{7})=\\=[; frac{6pi}{7}notin (-frac{pi}{2},frac{pi}{2}); ]=arcctg(tg(frac{pi}{2}+frac{5pi}{14}))=arcctg(-ctgfrac{5pi}{14})=\\=[; frac{5pi}{14}in (-frac{pi}{2},frac{pi}{2}); ,; arcctg(- alpha )=pi -arcctg alpha ; ]=\\=pi -arcctg(ctgfrac{5pi}{14})=pi -frac{5pi}{14}=frac{9pi}{14}$

$2)qquad arctg(tgx)=x; ,; -frac{pi}{2} textless x textless frac{pi}{2}\\arctg(ctg(-frac{21pi}{5}))=arctg(-ctg(4pi +frac{pi}{5}))=\\=[; arctg(- alpha )=-arctg alpha ; ]= -arctg(ctgfrac{pi}{5})=\\= -arctg(ctg(frac{pi}{2}-frac{3pi}{10}))= -arctg(tgfrac{3pi}{10})=\\=[; frac{3pi}{10}in (-frac{pi}{2},frac{pi}{2}); ]=-frac{3pi}{10}$